"The poet only asks to get his head into the heavens. It is the logician who seeks to get the heavens into his head. And it is his head that splits." G.K. Chesterton

Sunday, April 24, 2011

Number Theory

This result is a consequence of some musings in the last week. A few basic results will need to be proven first. The main result is:

Example. The above result guarantees for any x∊ ℤ a solution such that y ∊ ℤ in the following equations:
y = ½ ( ax3 + bx12 + cx81 + dx739 + ex993 - (a + b + c + d + e)x )
y = ½ ( a1x1 + a2x2 + a3x3 + ... + anx179 - (Σai)x )

Or more generally we have the following corollary for polynomials with integer terms:

I scribbled this Proposition (below) last week and wondered if it’s true in general. The proof turned out rather simple. I like this fact because of its general nature.

Proposition 1   For n∊ℤ+, k,x∊ℤ, and m∊ℕ.
xn + xn+m = 2k

The proof proceeds by showing first that given some integer of some parity, its powers remain of the same parity. For even, it follows immediately, for odd a simple argument using the Binomial Theorem shows that the remaining constant in an arbitrary expansion of 2k + 1 is 1, and the remining terms factorise as multiples of 2  i.e. (2k + 1)n =  2m +1 m∊ℤ.

The result from the top of the page is a direct corollary of Proposition 1.

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